
\prob{00A7}{Fourier级数}

求证：可积函数$f$可以表示为如下形式：
\[ f(t) = \frac{a_0}2 + \sum_{n = 1}^\infty (a_n\cos(n\omega t) + b_n\sin(n\omega t)) \]
其中
\begin{align*}
  a_n &= \frac1\pi \int_{-\pi}^\pi \cos(n\omega t)f(t) \dif t \\
  b_n &= \frac1\pi \int_{-\pi}^\pi \sin(n\omega t)f(t) \dif t
\end{align*}
\problabels{yellow/微积分, green/证明题}

\subsection{待定系数}

\begin{lemma} \label{lemma:00A7-tri}
  对于两不同正整数$n, k$，有
  \begin{align*}
    \int_{-\pi}^\pi \sin nx \dif x = \int_{-\pi}^\pi \cos nx \dif x &= 0 \\
    \int_{-\pi}^\pi \sin nx \cos nx \dif x &= 0 \\
    \int_{-\pi}^\pi \sin nx \cos kx \dif x &= 0 \\
    \int_{-\pi}^\pi \sin nx \sin kx \dif x &= 0 \\
    \int_{-\pi}^\pi \cos nx \cos kx \dif x &= 0
  \end{align*}
\end{lemma}

\begin{proof}
  参见总第~\ref{sec:00A6} 题。
\end{proof}

不妨设
\begin{align}
  f(t) = \frac{a_0}2 + \sum_{n = 1}^\infty (a_n\cos(n\omega t) + b_n\sin(n\omega t)) \label{eq:00A7-set}
\end{align}
再解出$a_n, b_n$。

对式~\ref{eq:00A7-set} 两边求定积分，得
\begin{align*}
  & \int_{-\pi}^\pi f(t) \dif t \\
  ={}& \int_{-\pi}^\pi \frac{a_0}2 \dif t \\
  &+ \int_{-\pi}^\pi \sum_{n = 1}^\infty (a_n\cos(n\omega t) + b_n\sin(n\omega t)) \dif t \\
  ={}& \int_{-\pi}^\pi \frac{a_0}2 \dif t + 0 = \left.\frac{a_0}2\right|_{-\pi}^\pi = \pi a_0 \\
\end{align*}
解得
\[ a_0 = \frac1\pi\int_{-\pi}^\pi f(t) \dif t \]

对于某正整数$k$，将式~\ref{eq:00A7-set} 两边同时乘以$\cos(k\omega t)$得
\begin{align*}
  & f(t)\cos(k\omega t) \\
  ={}& \frac{a_0}2\cos(k\omega t) + \sum_{n = 1}^\infty (a_n\cos(n\omega t)\cos(k\omega t) \\
  &+ b_n\sin(n\omega t)\cos(k\omega t))
\end{align*}
对等号两边求定积分，得
\begin{align*}
  & \int_{-\pi}^\pi f(t)\cos(k\omega t) \dif t \\
  ={}& \frac{a_0}2\int_{-\pi}^\pi \cos(k\omega t) \dif t \\
  &+ \sum_{n = 1}^\infty \left(a_n\int_{-\pi}^\pi \cos(n\omega t)\cos(k\omega t) \dif t\right. \\
  &+ \left.b_n\int_{-\pi}^\pi \sin(n\omega t)\cos(k\omega t) \dif t\right)
\end{align*}
由引理~\ref{lemma:00A7-tri} 可知，
\begin{align*}
  \int_{-\pi}^\pi \cos(k\omega t) \dif t &= 0 \\
  \int_{-\pi}^\pi \sin(n\omega t)\cos(k\omega t) \dif t &= 0
\end{align*}
且当$n \ne k$时，
\[ \int_{-\pi}^\pi \cos(n\omega t)\cos(k\omega t) \dif t = 0 \]
于是
\begin{align*}
  & \int_{-\pi}^\pi f(t)\cos(k\omega t) \dif t \\
  ={}& \sum_{n = 1}^\infty a_n\int_{-\pi}^\pi \cos(n\omega t)\cos(k\omega t) \dif t \\
  ={}& a_k\int_{-\pi}^\pi \cos^2(k\omega t) \dif t
\end{align*}
而
\begin{align*}
  \cos^2(k\omega t) &= \frac12\left(\cos^2(k\omega t) + \cos^2(k\omega t)\right) \\
  &= \frac12\left(1 + \cos^2(k\omega t) - \sin^2(k\omega t)\right) \\
  &= \frac12\left(1 + \cos(2k\omega t)\right)
\end{align*}
故
\begin{align*}
  & \int_{-\pi}^\pi f(t)\cos(k\omega t) \dif t \\
  ={}& a_k\int_{-\pi}^\pi (1 + \cos(2k\omega t)) \dif t \\
  ={}& a_k\left(\int_{-\pi}^\pi 1 \dif t + \int_{-\pi}^\pi \cos(2k\omega t) \dif t\right) \\
  ={}& 2\pi a_k + 0 = 2\pi a_k \\
\end{align*}
令$n = k$，解得
\[ a_n = \frac1\pi \int_{-\pi}^\pi \cos(n\omega t)f(t) \dif t \]
同理，将式~\ref{eq:00A7-set} 两边同时乘以$\sin(k\omega t)$得
\[ b_n = \frac1\pi \int_{-\pi}^\pi \sin(n\omega t)f(t) \dif t \]
经比较，与题中公式相同。证毕。
